Strings

Intuition Map characters letters, and map seen letters. Approach We will need to use two use two structures here. One to map the correspoding character One to keep track of seen chracters, this is necessary becase “no two chracter may map to the same character” If len is diff, just return false. Loop through the chracters (s -> cs, t -> ct). if cs was mapped before, and map[cs] == ct, move to the next if cs was mapped before, and map[cs] != ct, return False if cs was not mapped before, and seen[ct], return False else map[cs] = ct, and seen[ct] = True Complexity Time complexity: O(n) ...